Решите систему. Номер 437

Решение:
$\left \{ {{(x-2)(y+3)=160} \atop {y-x=1}} \right.\\ \left \{ {{(x-2)(y+3)=160} \atop {y=1+x}} \right. \\(x-2)(1+x+3)=160\\4x+ x^{2} -8-2x=160\\2x+-8-160=0\\ x^{2} +2x-168=0\\D1=1+168=169\\x1=12\\x2=-14\\y1=13\\y2=-13\\Answer: (12;13), (-14;-13)\\\\ \left \{ {{(x-1)(y+10)=9} \atop {x-y=11}} \right. \\ \left \{ {{(x-1)(y+10)=9} \atop {x=11+y}} \right. \\(11+y-1)(y+10)=9\\(10+y)(10+y)=9\\10+y=3\\y1=-7\\y2=-13\\x1=4\\x2=-2\\Answer: (4;-7), (-2;-13)$
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